Chapter 6: Systems of Equations 39

WRITING EQUATIONS 6.1.1 through 6.1.3

An equation is a mathematical sentence that conveys information to the reader. It uses variables and operation symbols (like +, - , /, =) to represent relationships for a situation. Technology has made it easier to solve equations, but to use these algebraic operating systems, you need to be able to write an equation. An equation may contain one or more variables. Solving a one variable equation (refer to sections 3.2, 5.1 and 5.2 in the textbook) involves different procedures than solving two variable equations. Students are asked to practice writing both types of equations in Chapter 6. See the Math Notes box on page 234.

Example 1 Just as there are variations in the vocabulary used in English sentences, there may be alternative ways to write an equation. The sum of the lengths of two adjacent sides of a rectangle is 15. Using one variable: If x = the length, then the width = 15! x , so x + (15 ! x) = 15 Using two variables: l + w = 15 , where l is the length and w is the width. Example 2 Equations are easier to use if the variable is defined for the reader. a) A collection of quarters, dimes and

nickels is worth $6.00. Let q = quarters, d = dimes, and n = nickels.

0.25q + 0.10d + 0.05n = 6.00

b) The cost of cell phone calls priced at 5¢ per minute. Letc = cost, m = minutes.

c = 0.05m

Algebra Connections Parent Guide 40

Example 3 Drawing a diagram with labels or creating a Guess & Check table can help identify relationships between quantities. The perimeter of a rectangle is 60 cm. The width is 4 times the length. Write an equation and solve for the length. Guess & Check Table:

Length (guess) Width Perimeter Check? ? 10 5

4(10) 4(5)

2(10) + 2(40) 2(5) + 2(20)

100 50

too big too small

l 4(l) 2l + 2(4l) = 60 Drawing a diagram: Using one variable: 2l + 2(4l) = 60

Using two variables and two equations: w = 4l

2l + 2w = 60

Example 4 A “Let” statement describes what the variable represents. Mike spent $11.19 on a bag containing red and blue candies. The bag weighed 11 pounds. The red candy costs $1.29 a pound and the blue candy costs $0.79 a pound. Write a system of equations for this problem. Let x = the weight of the red candy and y = the weight of the blue candy.

x + y = 11

1.29x + 0.79y = 11.19

The first equation represents the weight of the bag. The second equation represents the cost of the candy.

4l

l or

w

l

Chapter 6: Systems of Equations 41

Problems Write an equation using one variable or a system of equations. Do NOT solve your equations. 1. A rectangle is three times as long as it is wide. Its perimeter is 36. Find each side length. 2. A rectangle is twice as long as it is wide. Its area is 72. Find each side length. 3. The sum of two consecutive odd integers is 76. Find the numbers. 4. Nancy started the year with $425 in the bank and is saving $25 a week. Seamus started with

$875 and is spending $15 a week. When will they both have the same amount of money in the bank?

5. Oliver earns $50 a day and $7.50 for each package he processes at Company A. His

paycheck on his first day was $140. How many packages did he process? 6. Dustin has a collection of quarters and pennies. The total value is $4.65. There are 33

coins. How many quarters and pennies does he have? 7. A one pound mixture of raisins and peanuts cost $7.50. The raisins cost $3.25 a pound and

the peanuts cost $5.75 a pound. How much of each ingredient is in the mixture? 8. An adult ticket at an amusement park costs $24.95 and a child’s ticket costs $15.95. A

group of 10 people paid $186.50 to enter the park. How many were adults? 9. Katy weighs 105 pounds and is gaining 2 pounds a month. James weighs 175 pounds and is

losing 3 pounds a month. When will they weigh the same? 10. Holmes Junior High has x students. Harper Middle School has 125 fewer students than

Holmes. When the two schools are merged there will be 809 students. How many students attend each school?

Algebra Connections Parent Guide 42

Answers (Other equivalent forms are possible.)

One Variable Equation System of Equations Let Statement

1.

2w + 2(3w) = 36

w = 3l

2w + 2l = 36 Let l = length, w = width

2. w(2w) = 72

w = 2l

lw = 72 Let l = length, w = width

3. m + (m + 2) = 76 m + n = 76

n = m + 2 Letm = the first odd integer

and n = the next consecutive odd integer

4. 425 + 25x = 875 !15x y = 425 + 25x

y = 875 !15x

Let x = the number of weeks and y = the total money in the bank

5. 50 + 7.5p = 140

6. 0.25q + 0.01(33! q) = 4.65

q + p = 33

0.25q + 0.01p = 4.65

Letq = number of quarters, p = number of pennies

7. 3.25r + 5.75(1! r) = 7.5

r + p = 1

3.25r + 5.75p = 7.5(1)

Let r = weight of raisins and p = weight of peanuts

8. 24.95a +15.95(10 ! a) = 186.5 a + c = 10

24.95a +15.95c = 186.5 Let a = number of adult

tickets and c = number of child’s tickets

9. 105 + 2m = 175 ! 3m w = 105 + 2m

w = 175 ! 3m

Letm = the number of months and w = the weight of each person

10 x + (x !125) = 809 x + y = 809

y = x !125

Let x = Holmes students and y = Harper students

Chapter 6: Systems of Equations 43

SOLVING SYSTEMS BY SUBSTITUTION 6.2.1

A system of equations has two or more equations with two or more variables. In Section 4.2, students were introduced to solving a system by looking at the intersection point of the graphs. They also learned the algebraic “equal values” solution method. Graphing and the Equal Values method are convenient when both equations are in y-form. The substitution method is used to change a two-variable system of equations to a one-variable equation. This method is useful when one of the variables is isolated in one of the equations in the system. Two substitutions must be made to find both the x- and y-values for a complete solution. See the Math Notes boxes on pages 248, 252, and 258.

The equation

x + y = 9 has infinite possibilities for solutions: (10, -1) (2, 7) (0, 9)…, but if

y = 4 then there is only one possible value for x. That value is easily seen when we replace (substitute for) y with 4 in the original equation: x + 4 = 9 , so x = 5when y = 4 . Substitution and this observation are the basis for the following method to solve systems of equations. Example 1 When solving a system of equations, we replace a variable with an algebraic expression.

Solve: y = 2x

x + y = 9

x + y = 9 First replace y with 2x so there is only one variable in the

second equation.

x + 2x = 9

3x = 9

x = 3

Solve as usual.

y = 2(3) Use x = 3 to substitute in either original equation to find y = 6 the y value.

The solution is (3, 6) and makes both equations true. When graphing, this is the intersection point of the two lines.

Algebra Connections Parent Guide 44

Example 2 Substitute the variable that is convenient. Look for the variable that is by itself on one side of the equation.

Solve: 4x + y = 8

x = 5 ! y

4(5 ! y) + y = 8

20 ! 4y + y = 8

20 ! 3y = 8

!3y = !12

y = 4

x = 5 ! 4

x = 1

Since x is alone, substitute 5 ! y in the first equation for x, then solve as usual. Then substitute y = 4 in the original equation to find x.

The solution is (1, 4). Check by substituting this point into each equation to confirm that you are correct. Example 3 Not all systems have a solution. If the substitution results in an equation that is not true, then there is NO SOLUTION. This graph would have two parallel lines, so there is no point of intersection.

Solve: y = 7 ! 3x

3x + y = 10

3x + (7 ! 3x) = 10

3x ! 3x + 7 = 10

7 " 10

Substitute 7 ! 3x in the second equation for y. No Solution.

There may also be an INFINITE NUMBER OF SOLUTIONS. This graph would appear as a single line for the two equations.

Solve: y = 4 ! 2x

!4x ! 2y = !8

!4x ! 2(4 ! 2x) = !8

!!!!!4x ! 8 + 4x = !8

!8 = !8

Substitute 4 ! 2x in the second equation for y. Always true for any value of x or y.

Chapter 6: Systems of Equations 45

Example 4 Sometimes you need to solve one of the equations for x or y to use this method. Also be careful when a negative sign appears in the equation. Check your final solution to be sure it works in both of the original equations.

Solve:

y ! 2x = !7

3x ! 4y = 8

3x ! 4(2x ! 7) = 8

3x ! 8x + 28 = 8

!5x + 28 = 8

!5x = !20

x = 4

Solve the first equation for y to get y = 2x ! 7 , then substitute 2x ! 7 in the second equation for y.

Then substitute again to find y.

y = 2 4( ) ! 7

y = 1

Check: 1! 2(4) = !7 and 3(4) ! 4(1) = 8 .

The solution is (4, 1). This would be the intersection point of the two lines. Problems 1. y = !3x

4x + y = 2

2. y = 7x ! 5

2x + y = 13

3. x = !5y ! 4

x ! 4y = 23

4. x + y = 10

y = x ! 4

5. y = 5 ! x

4x + 2y = 10

6. 3x + 5y = 23

y = x + 3

7. y = !x ! 2

2x + 3y = !9

8. y = 2x ! 3

!2x + y = 1

9. x = 1

2y + 1

2

2x + y = !1

10. a = 2b + 4

b ! 2a = 16 11. y = 3! 2x

4x + 2y = 6

12. y = x +1

x ! y = 1

Answers 1. (2, -6) 2. (2, 9) 3. (11, -3)

4. (7, 3) 5. (0, 5) 6. (1, 4)

7. (3, -5) 8. No solution 9. (0, -1)

10. (-12, -8) 11. Infinite solutions 12. No solution

Algebra Connections Parent Guide 46

ELIMINATION METHOD 6.2.3 and 6.2.4

In previous work with systems of equations, sections 6.1 and 6.2, one of the variables was usually alone on one side of one of the equations. In that situation, you can substitute to rewrite a two-variable equation as a one-variable equation, then solve the resulting equation. Systems of linear equations in standard form (that is, ax + by = c ) have both variables on the same side of both equations. To solve this type of system, we can eliminate one of the variables by focusing on the coefficients. (The coefficient is the number in front of the variable.) This method is called the elimination method and uses addition or subtraction, sometimes combined with multiplication or division, to rewrite the system as a one variable equation. After solving for one variable, we then substitute to find the other. A complete solution includes both an x and a y value. See the Math Notes boxes on pages 264 and 287 as well as those on pages 252 and 258.

Example 1

Solve: x ! y = 2

2x + y = 1 Notice that y and – y are opposites.

Adding the two equations together will eliminate the y-term because y + !y = 0 . This is sometimes called the addition method of elimination.

Add vertically. x ! y = 2

+2x + y = 1

Solve for x. 3x + 0 = 3

x = 1

Substitute the value of x in either of the original equations to find the y value.

2(1) + y = 1

2 + y = 1

y = !1

The point of intersection is (1, -1).

Chapter 6: Systems of Equations 47

Example 2

Solve: 3x + 6y = 24

3x + y = !1 Notice that 3x ! 3x = 0 .

If the coefficients are the same, subtract the second equation to eliminate the variable, then substitute the value of y into one of the original equations to find x.

3x + 6y = 24

!(3x + y = !1)

3x + 6y = 24

!3x ! y = +1

0 + 5y = 25

y = 5

3x + 6(5) = 24

3x + 30 = 24

3x = !6

x = !2

The solution is (-2, 5). It makes both equations true and names the point where the two lines intersect on a graph. Example 3 If neither of the pairs of coefficients are the same, multiply one or both of the equations to make one pair of coefficients the same or opposites. Then add or subtract as needed to eliminate a variable term.

Solve: x + 3y = 7

4x ! 7y = !10

Multiplying the first equation by 4 or -4 will allow you to eliminate the x. Be sure to multiply everything on the equation’s left side and right side to maintain equivalence.

One possible process: multiply the first equation by 4. 4(x + 3y = 7)! 4x +12y = 28 4x +12y = 28

!(4x ! 7y = !10)

19y = 38

y = 2

4x ! 7(2) = !10

4x !14 = !10

4x = 4

x = 1

Use this equation in the system with the original second equation. Subtract and solve for y, then substitute in either original equation to find x. Solution is (1, 2).

If you prefer the addition method, multiply the first equation by – 4:

!4(x + 3y = 7)"!4x !12y = !28 . Then add and solve.

then

Algebra Connections Parent Guide 48

Example 4 If one multiplication will not eliminate a variable, multiply both equations by different numbers to get coefficients that are the same or opposites.

Solve: 8x ! 7y = 5

3x ! 5y = 9

Two ways to eliminate x:

3(8x ! 7y = 5)" 24x ! 21y = 15

8(3x ! 5y = 9)" 24x ! 40y = 72 or

3(8x ! 7y = 5)" 24x ! 21y = 15

!8(3x ! 5y = 9)"!24x + 40y = !72

Two ways to eliminate y:

5(8x ! 7y = 5)" 40x ! 35y = 25

7(3x ! 5y = 9)" 21x ! 35y = 63 or

!5(8x ! 7y = 5)"!40x + 35y = !25

7(3x ! 5y = 9)" 21x ! 35y = 63

The examples above are all equivalent systems and have the same solution of (-2, -3). Example 5 The special cases of no solution and infinite solutions can also occur. See the Math Notes box on page 258.

Solve: 4x + 2y = 6

2x + y = 3 Dividing the first equation by 2 produces 2x + y = 3 .

The two equations are identical, so when graphed there would be one line with infinite solutions because the same ordered pairs are true for both equations.

Solve: 2x + y = 3

4x + 2y = 8 Multiplying the first equation by 2 produces 4x + 2y = 6 .

There are no numbers that could make both equations true at the same time, so the lines are parallel and there is no solution, that is, no point of intersection.

Chapter 6: Systems of Equations 49

SUMMARY OF METHODS TO SOLVE SYSTEMS

Method When to Use Example Equal values Both equations in y-form.

y = x ! 2

y = !2x +1

Substitution One variable is alone on one side of one equation.

y = !3x !1

3x + 6y = 24

Add to eliminate Equations in standard form with opposite coefficients.

x + 2y = 21

3x ! 2y = 7

Subtract to eliminate Equations in standard form with same coefficients.

x + 2y = 3

3x + 2y = 7

Multiply to eliminate When nothing else works. In this case you could multiply the first equation by 3 and the second equation by 2, then subtract to eliminate x.

2x ! 5y = 3

3x + 2y = 7

Problems 1. 2x + y = 6

!2x + y = 2

2. !4x + 5y = 0

!6x + 5y = !10

3. 2x ! 3y = !9

x + y = !2

4. y ! x = 4

2y + x = 8

5. 2x ! y = 4

1

2x + y = 1

6. !4x + 6y = !20

2x ! 3y = 10

7. 6x ! 2y = !16

4x + y = 1

8. 6x ! y = 4

6x + 3y = !16

9. 2x ! 2y = 5

2x ! 3y = 3

10. y ! 2x = 6

y ! 2x = !4

11. 4x ! 4y = 14

2x ! 4y = 8

12. 3x + 2y = 12

5x ! 3y = !37

Answers 1. (1, 4)

2. (5, 4) 3. (!3 , 1)

4. (0, 4)

5. (2, 0) 6. Infinite

7. (!1, 5)

8. ( !

1

6, !5 ) 9. (4.5, 2)

10. No solution 11. (3, !

1

2) 12. (!2 , 9)